Problem: What is the value of $\dfrac{d}{dx}\left(-\dfrac{1}{x^2}-\dfrac{7}{x}\right)$ at $x=-2$ ?
Explanation: The strategy We can first rewrite each rational term in the expression as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Once we have the derivative, we can evaluate it at $x=-2$. Rewriting rational terms as negative powers $\begin{aligned} &\phantom{=}-\dfrac{1}{x^2}-\dfrac{7}{x} \\\\ &=-x^{-2}-7x^{-1} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}(-x^{-2}-7x^{-1} ) \\\\ &=-1\dfrac{d}{dx}(x^{-2})-7\dfrac{d}{dx}(x^{-1}) \\\\ &=-1(-2x^{-3})-7(-1)x^{-2} \\\\ &=2x^{-3}+7x^{-2} \\\\ &=\dfrac{2}{x^3}+\dfrac{7}{x^2} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(-\dfrac{1}{x^2}-\dfrac{7}{x}\right)=\dfrac{2}{x^3}+\dfrac{7}{x^2}$. Let's evaluate it at $x=-2$ : $\begin{aligned} &\phantom{=}\dfrac{2}{x^3}+\dfrac{7}{x^2} \\\\ &=\dfrac{2}{(-2)^3}+\dfrac{7}{(-2)^2} \\\\ &=-\dfrac{2}{8}+\dfrac{7}{4} \\\\ &=-\dfrac{1}{4}+\dfrac{7}{4} \\\\ &=\dfrac{6}{4} \\\\ &=\dfrac{3}{2} \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(-\dfrac{1}{x^2}-\dfrac{7}{x}\right)$ at $x=-2$ is $\dfrac{3}{2}$.